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Thread: some modulus questions

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    Only Slightly Obsessive jemidiah's Avatar
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    Re: some modulus questions

    If you showed us your current work I'd be happy to go over it, but I'm a little anxious about answering questions on a quiz you may or may not have submitted. Sorry!
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    Re: some modulus questions

    lol. okey.
    number 5 answer is D

    For 3, I got A as answer.

    For the other questions, I just eliminated the denominator but a friend of mine is telling me to multiply both side by the (dominator)^2.
    Just eliminating the denominator seems too simple to do.
    I'm a little unsure about my friend's theory; the left hand side can be practically anything.

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    Only Slightly Obsessive jemidiah's Avatar
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    Re: some modulus questions

    3. A: Let x-> -1.5 and you'll see the counterexample immediately.
    4. -2 < x < -1. See 4* for the work/reasoning.
    5. D: Let x-> 2 and you'll see immediately that the other functions just don't work.
    6. Apply the reasoning from 4* and you'll get these two. Alternatively, you can just graph the functions (this is probably easier).


    4*. First off, the inequality doesn't work whenever the denominator is 0, which happens at x = 2, and x = -2. So, x cannot be 2, or -2.

    With that in mind, you can multiply both sides by the denominator, but you MUST remember that multiplying by a negative number flips the direction of the inequality.

    So, when is the denominator positive and when is it negative? When x > -2, the denominator is positive; negative otherwise (note that x <> +/- 2).

    Thus, for x > -2, the inequality becomes x+1 < 0 or x < -1. Both of these are satisfied only for -2 < x < -1

    Thus, for x < -2, the inequality becomes x+1 > 0 or x > -1. Both of these are satisfied, well, never.

    So the solution is just -2 < x < -1
    The time you enjoy wasting is not wasted time.
    Bertrand Russell

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